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In the derivation of the LSZ reduction formula equation (5.21) Srednicki claims that in case of an interaction term in the Lagrangian density $a^{\dagger}(\textbf{k})$ will no longer be time dependent. Now does that mean that the only modification to the formula of $\varphi(x)$ will be a time dependence of $a$'s and $a^{\dagger}$'s? So is that like an ansatz we use for $\varphi(x)$? Now even if we do assume this form of $\varphi(x)$, in line two of the derivation he makes the substitution $$a^{\dagger}(\textbf{k},t)= -i \int\:d^3x \:[e^{ikx} \partial_0\varphi(x) - e^{ikx} ik_0 \varphi(x)]\tag{5.2}.$$ Now in the derivation of this formula if we remember we explicitly had to use the time independence of the $a$'s. So what am I missing in this derivation that justifies the substitution in the case the $a$'s are time dependent?

Qmechanic
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Soumil
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    I think $a^\dagger(k,t) = ...$ might have to be thought of as a definition? Maybe look at the answer here https://physics.stackexchange.com/a/404321/157704 – user1379857 Apr 06 '19 at 21:03
  • Possible duplicate: https://physics.stackexchange.com/q/525800/2451 – Qmechanic Jan 25 '20 at 20:57

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That expression is time independent. Try taking the time derivative, being sure to use the equation of motion for $\varphi(x)$, also notice that the plane wave $e^{ik\cdot x}$ satisfies the same equation of motion.

fewfew4
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    Yes, it is time independent for the free field theory for which we can use $(-\partial^2 + m^2) \varphi(x)=0 $. But I'm asking how is it justified to use the same expression for $a^{\dagger}(t)$ in the interacting theory. In the later steps he understandably doesn't use the previous equation of motion for the free field theory. Now both of them match at $t$ goes to $\pm \infty$. But is that enough to extrapolate(backwards) that at general times for the free field theory we can use this as the definition of $a^{\dagger}(t)$ ? – Soumil Apr 07 '19 at 20:32